[足式机器人]Part4 南科大高等机器人控制课 Ch03 Operator View of Rigid-Body Transformation

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B站：CLEAR_LAB
笔者带更新-运动学
课程主讲教师：
Prof. Wei Zhang

南科大高等机器人控制课 Ch03 Operator View of Rigid-Body Transformation

[1. Rotation Operation via Differential Equation](#1. Rotation Operation via Differential Equation)
[1.1 Skew Symmetric Matrices](#1.1 Skew Symmetric Matrices)
- [1.2 Rotation Operation via Differential Equation](#1.2 Rotation Operation via Differential Equation)
[1.3 Raotation Matrix as a Rotation Operator](#1.3 Raotation Matrix as a Rotation Operator)
[2. Rotation Operation in Different Frames](#2. Rotation Operation in Different Frames)
- [2.1 Rotation Martix Properties](#2.1 Rotation Martix Properties)
- [2.2 Rotation Operation in Different Frames](#2.2 Rotation Operation in Different Frames)
[3. Rigid-Body Operation via Diffeential Equation](#3. Rigid-Body Operation via Diffeential Equation)
- [3.1 SE(3)](#3.1 SE(3))
- [3.2 se(3)](#3.2 se(3))
[4. Homogeneous Transformation Matrix as Rigid-Body Operator](#4. Homogeneous Transformation Matrix as Rigid-Body Operator)
[5. Rigid-Body Operation of Screw Axis](#5. Rigid-Body Operation of Screw Axis)

1. Rotation Operation via Differential Equation

1.1 Skew Symmetric Matrices

Recall that cross product is a special linear transformation.

For any ω ⃗ ∈ R 3 \vec{\omega}\in \mathbb{R} ^3 ω ∈R3, there is a matrix ω ⃗ ~ ∈ R 3 × 3 \tilde{\vec{\omega}}\in \mathbb{R} ^{3\times 3} ω ~∈R3×3 such that ω ⃗ × R ⃗ = ω ⃗ ~ R ⃗ \vec{\omega}\times \vec{R}=\tilde{\vec{\omega}}\vec{R} ω ×R =ω ~R
ω ⃗ = $ω 1 ω 2 ω 3$ ⟷ ω ⃗ ~ = $0 - ω 3 ω 2 ω 3 0 - ω 1 - ω 2 ω 1 0$ \vec{\omega}=\left $\\begin{array}{c} \\omega _1\\\\ \\omega _2\\\\ \\omega _3\\\\ \\end{array} \\right$ \longleftrightarrow \tilde{\vec{\omega}}=\left $\\begin{matrix} 0\& -\\omega _3\& \\omega _2\\\\ \\omega _3\& 0\& -\\omega _1\\\\ -\\omega _2\& \\omega _1\& 0\\\\ \\end{matrix} \\right$ ω = ω1ω2ω3 ⟷ω ~= 0ω3−ω2−ω30ω1ω2−ω10

Note that a ⃗ ~ = − a ⃗ ~ T \tilde{\vec{a}}=-\tilde{\vec{a}}^{\mathrm{T}} a ~=−a ~T (called skew stmmetric )
ω ⃗ ~ \tilde{\vec{\omega}} ω ~ is called a skew-symmetric matrix representation of the vector ω ⃗ \vec{\omega} ω

The set of skew-symmetric matrices in : s o ( n ) ≜ { S ∈ R n × n : S T = − S } so\left( n \right) \triangleq \left\{ S\in \mathbb{R} ^{n\times n}:S^{\mathrm{T}}=-S \right\} so(n)≜{S∈Rn×n:ST=−S}

We are interested in case n = 2 , 3 n=2,3 n=2,3

Rotation matrix ∈ S O ( 3 ) { R T R = E , det ⁡ ( R ) = 1 } \in SO\left( 3 \right) \left\{ R^{\mathrm{T}}R=E,\det \left( R \right) =1 \right\} ∈SO(3){RTR=E,det(R)=1}

1.2 Rotation Operation via Differential Equation

Consider a point initially located at R ⃗ p 0 \vec{R}_{p_0} R p0 at time t = 0 t=0 t=0
Rotate the point with unit angular velocity ω ^ \hat{\omega} ω^. Assuming the rotation axis passing through the origin, the motion is describe by
R ⃗ ˙ p ( t ) = ω ^ × R ⃗ p = ω ^ ~ R ⃗ p ( t ) , p ( 0 ) = p 0 \dot{\vec{R}}_p\left( t \right) =\hat{\omega}\times \vec{R}_p=\tilde{\hat{\omega}}\vec{R}_p\left( t \right) ,p\left( 0 \right) =p_0 R ˙p(t)=ω^×R p=ω^~R p(t),p(0)=p0
linear velocity at time t t t, recall x ˙ = A x , x ( 0 ) = x 0 ⇒ x ( t ) = e A t x 0 \dot{x}=Ax,x\left( 0 \right) =x_0\Rightarrow x\left( t \right) =e^{At}x_0 x˙=Ax,x(0)=x0⇒x(t)=eAtx0
This is a linear ODE with solution : R ⃗ ˙ p ( t ) = ω ^ ~ R ⃗ p ( t ) ⇒ R ⃗ ˙ p ( t ) = e ω ^ ~ t R ⃗ p 0 \dot{\vec{R}}_p\left( t \right) =\tilde{\hat{\omega}}\vec{R}_p\left( t \right) \Rightarrow \dot{\vec{R}}p\left( t \right) =e^{\tilde{\hat{\omega}}t}\vec{R}{p_0} R ˙p(t)=ω^~R p(t)⇒R ˙p(t)=eω^~tR p0
After t = θ t=\theta t=θ, the point has been rotated by θ \theta θ degree. Note R ⃗ ˙ p ( θ ) = e ω ^ ~ θ R ⃗ p 0 \dot{\vec{R}}p\left( \theta \right) =e^{\tilde{\hat{\omega}}\theta}\vec{R}{p_0} R ˙p(θ)=eω^~θR p0, e ω ^ ~ θ e^{\tilde{\hat{\omega}}\theta} eω^~θ is rotation operator
R o t ( ω ^ , θ ) ≜ e ω ^ ~ θ \mathrm{Rot}\left( \hat{\omega},\theta \right) \triangleq e^{\tilde{\hat{\omega}}\theta} Rot(ω^,θ)≜eω^~θ can be viewed as a rotation operator that rotates a point about ω ^ \hat{\omega} ω^ through θ \theta θ degree

The discussion holds for any reference frame

1.3 Raotation Matrix as a Rotation Operator

Theorem: Every rotation matrix R R R can be written as R = R o t ( ω ^ , θ ) ≜ e ω ^ ~ θ ∈ S O ( 3 ) { R T R = E , det ⁡ ( R ) = 1 } R=\mathrm{Rot}\left( \hat{\omega},\theta \right) \triangleq e^{\tilde{\hat{\omega}}\theta}\in SO\left( 3 \right) \left\{ R^{\mathrm{T}}R=E,\det \left( R \right) =1 \right\} R=Rot(ω^,θ)≜eω^~θ∈SO(3){RTR=E,det(R)=1}, i.e. , it represents a rotation operation about ω ^ \hat{\omega} ω^ by θ \theta θ
We have seen how to use R R R to represent frame orientation and change of coordinate between different frames. They are quite different from the operator interpretation of R R R
To apply the rotation operation, all the vectors / matrices have to be expressed in the same refrence frame

For example, assume R = $1 0 0 0 0 - 1 0 1 0$ = R o t ( x ^ , π 2 ) R=\left $\\begin{matrix} 1\& 0\& 0\\\\ 0\& 0\& -1\\\\ 0\& 1\& 0\\\\ \\end{matrix} \\right$ =\mathrm{Rot}\left( \hat{x},\frac{\pi}{2} \right) R= 1000010−10 =Rot(x^,2π)

Consider a relation q = R p q=Rp q=Rp:

Change reference frame interpretation : two frames { A } , { B } \left\{ A \right\} ,\left\{ B \right\} {A},{B} , one physical Point a a a
R R R : orientation of { B } \left\{ B \right\} {B} relative to { A } \left\{ A \right\} {A} : i.e. R = $Q B A$ R=\left $Q_{\\mathrm{B}}\^{A} \\right$ R= $QBA$
then : p = a B , q = a A , q = R p ⇒ a A = $Q B A$ a B p=a^B,q=a^A,q=Rp\Rightarrow a^A=\left $Q_{\\mathrm{B}}\^{A} \\right$ a^B p=aB,q=aA,q=Rp⇒aA= $QBA$ aB

Rotation operator interpretation : one frame and two points
a → a ′ , p = a A , q = a ′ A ⇒ a ′ A = R a A a\rightarrow a^{\prime},p=a^A,q={a^{\prime}}^A\Rightarrow {a^{\prime}}^A=Ra^A a→a′,p=aA,q=a′A⇒a′A=RaA

Consider the frame operation:

Change of reference frame :
Have one "frame object", two reference frames
Frame object { A } \left\{ A \right\} {A}, orientation in { O } \left\{ O \right\} {O} , is R A O , R A B R_{\mathrm{A}}^{O},R_{\mathrm{A}}^{B} RAO,RAB, ⇒ R A O = R B O R A B = R R A B \Rightarrow R_{\mathrm{A}}^{O}=R_{\mathrm{B}}^{O}R_{\mathrm{A}}^{B}=RR_{\mathrm{A}}^{B} ⇒RAO=RBORAB=RRAB

Rotation a frame : R ′ A = R R A {R^{\prime}}A=RR_A R′A=RRA
two frame objects, one refrence frame
more precisely : R ′ A = R R A ⇒ R A ′ O = R R A O {R^{\prime}}A=RR_A\Rightarrow R{\mathrm{A}^{\prime}}^{O}=RR{\mathrm{A}}^{O} R′A=RRA⇒RA′O=RRAO

2. Rotation Operation in Different Frames

2.1 Rotation Martix Properties

$Q$ $Q$ T = E \left $Q \\right$ \left $Q \\right$ ^{\mathrm{T}}=E $Q$ $Q$ T=E

$Q 1$ $Q 2$ ∈ S O ( 3 ) , i f $Q 1$ , $Q 2$ ∈ S O ( 3 ) \left $Q_1 \\right$ \left $Q_2 \\right$ \in SO\left( 3 \right) ,if\,\,\left $Q_1 \\right$ ,\left $Q_2 \\right$ \in SO\left( 3 \right) $Q1$ $Q2$ ∈SO(3),if $Q1$ , $Q2$ ∈SO(3)： product of two rotation matrices is also a rotation matrix
p , q ∈ R 3 , ∥ $Q$ R ⃗ p − $Q$ R ⃗ q ∥ 2 = ∥ $Q$ ( R ⃗ p − R ⃗ q ) ∥ 2 = ( R ⃗ p − R ⃗ q ) T $Q$ T $Q$ ( R ⃗ p − R ⃗ q ) = ∥ R ⃗ p − R ⃗ q ∥ 2 p,q\in \mathbb{R} ^3,\left\| \left $Q \\right$ \vec{R}p-\left $Q \\right$ \vec{R}{\mathrm{q}} \right\| ^2=\left\| \left $Q \\right$ \left( \vec{R}p-\vec{R}{\mathrm{q}} \right) \right\| ^2=\left( \vec{R}p-\vec{R}{\mathrm{q}} \right) ^{\mathrm{T}}\left $Q \\right$ ^{\mathrm{T}}\left $Q \\right$ \left( \vec{R}p-\vec{R}{\mathrm{q}} \right) =\left\| \vec{R}p-\vec{R}{\mathrm{q}} \right\| ^2 p,q∈R3, $Q$ R p− $Q$ R q 2= $Q$ (R p−R q) 2=(R p−R q)T $Q$ T $Q$ (R p−R q)= R p−R q 2

∥ R ⃗ p − R ⃗ q ∥ = ∥ $Q$ R ⃗ p − $Q$ R ⃗ q ∥ \left\| \vec{R}p-\vec{R}{\mathrm{q}} \right\| =\left\| \left $Q \\right$ \vec{R}p-\left $Q \\right$ \vec{R}{\mathrm{q}} \right\| R p−R q = $Q$ R p− $Q$ R q : rotation operator preserves distance

$Q$ ( v ⃗ × w ⃗ ) = $Q$ v ⃗ × $Q$ w ⃗ \left $Q \\right$ \left( \vec{v}\times \vec{w} \right) =\left $Q \\right$ \vec{v}\times \left $Q \\right$ \vec{w} $Q$ (v ×w )= $Q$ v × $Q$ w

$Q$ w ⃗ ~ $Q$ T = $Q$ w ⃗ ~ \left $Q \\right$ \tilde{\vec{w}}\left $Q \\right$ ^{\mathrm{T}}=\widetilde{\left $Q \\right$ \vec{w}} $Q$ w ~ $Q$ T= $Q$ w ------ important

2.2 Rotation Operation in Different Frames

Consider two frames { A } \left\{ A \right\} {A} and { B } \left\{ B \right\} {B}， the actual numerical values of the operator R o t ( ω ^ , θ ) \mathrm{Rot}\left( \hat{\omega},\theta \right) Rot(ω^,θ) depend on both the reference frame to repersent ω ^ \hat{\omega} ω^ and the reference frame to represent the operator itself

Consider a rotation axis ω ^ \hat{\omega} ω^ (coordinate free vector), with { A } \left\{ A \right\} {A} - frame coordinate ω ^ A \hat{\omega}^A ω^A and { B } \left\{ B \right\} {B} - frame. We know ω ⃗ A = $Q B A$ ω ⃗ B \vec{\omega}^A=\left $Q_{\\mathrm{B}}\^{A} \\right$ \vec{\omega}^B ω A= $QBA$ ω B

Let B R o t ( B ω ^ , θ ) ^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) BRot(Bω^,θ) and A R o t ( A ω ^ , θ ) ^A\mathrm{Rot}\left( ^A\hat{\omega},\theta \right) ARot(Aω^,θ) be the two rotation matrices, representing the same rotation operatioon R o t ( ω ^ , θ ) \mathrm{Rot}\left( \hat{\omega},\theta \right) Rot(ω^,θ) in frames { A } \left\{ A \right\} {A} and { B } \left\{ B \right\} {B}

We have the relation : A R o t ( A ω ^ , θ ) = $Q B A$ B R o t ( B ω ^ , θ ) $Q A B$ ^A\mathrm{Rot}\left( ^A\hat{\omega},\theta \right) =\left $Q_{\\mathrm{B}}\^{A} \\right$ ^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) \left $Q_{\\mathrm{A}}\^{B} \\right$ ARot(Aω^,θ)= $QBA$ BRot(Bω^,θ) $QAB$

Approach 1 : two points p → p ′ ⇒ R ⃗ p ′ A = A R o t ( A ω ^ , θ ) R ⃗ p A p\rightarrow p^{\prime}\Rightarrow \vec{R}{\mathrm{p}^{\prime}}^{A}=^A\mathrm{Rot}\left( ^A\hat{\omega},\theta \right) \vec{R}{\mathrm{p}}^{A} p→p′⇒R p′A=ARot(Aω^,θ)R pA
{ B } \left\{ B \right\} {B} - frame : R ⃗ p ′ B = B R o t ( B ω ^ , θ ) R ⃗ p B \vec{R}{\mathrm{p}^{\prime}}^{B}=^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) \vec{R}{\mathrm{p}}^{B} R p′B=BRot(Bω^,θ)R pB
⇒ $Q B A$ R ⃗ p ′ B = $Q B A$ B R o t ( B ω ^ , θ ) R ⃗ p B ⇒ R ⃗ p ′ A = $Q B A$ B R o t ( B ω ^ , θ ) $Q B A$ R ⃗ p A ⇒ A R o t ( A ω ^ , θ ) = $Q B A$ B R o t ( B ω ^ , θ ) $Q B A$ \Rightarrow \left $Q_{\\mathrm{B}}\^{A} \\right$ \vec{R}{\mathrm{p}^{\prime}}^{B}=\left $Q_{\\mathrm{B}}\^{A} \\right$ ^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) \vec{R}{\mathrm{p}}^{B}\Rightarrow \vec{R}{\mathrm{p}^{\prime}}^{A}=\left $Q_{\\mathrm{B}}\^{A} \\right$ ^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) \left $Q_{\\mathrm{B}}\^{A} \\right$ \vec{R}{\mathrm{p}}^{A}\Rightarrow ^A\mathrm{Rot}\left( ^A\hat{\omega},\theta \right) =\left $Q_{\\mathrm{B}}\^{A} \\right$ ^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) \left $Q_{\\mathrm{B}}\^{A} \\right$ ⇒ $QBA$ R p′B= $QBA$ BRot(Bω^,θ)R pB⇒R p′A= $QBA$ BRot(Bω^,θ) $QBA$ R pA⇒ARot(Aω^,θ)= $QBA$ BRot(Bω^,θ) $QBA$
Approach 2 : for a ⃗ ∈ R 3 , a ⃗ ~ ∈ s o ( 3 ) , $Q$ ∈ S O ( 3 ) \vec{a}\in \mathbb{R} ^3,\tilde{\vec{a}}\in so\left( 3 \right) ,\left $Q \\right$ \in SO\left( 3 \right) a ∈R3,a ~∈so(3), $Q$ ∈SO(3)
⇒ $Q$ a ⃗ ∈ R 3 , $Q$ a ⃗ ~ = $Q$ a ⃗ ~ $Q$ T \Rightarrow \left $Q \\right$ \vec{a}\in \mathbb{R} ^3,\widetilde{\left $Q \\right$ \vec{a}}=\left $Q \\right$ \tilde{\vec{a}}\left $Q \\right$ ^{\mathrm{T}} ⇒ $Q$ a ∈R3, $Q$ a = $Q$ a ~ $Q$ T
R o t ( ω ⃗ A , θ ) = e ω ⃗ ~ A θ = e $Q B A$ ω ⃗ B ~ θ = e $Q B A$ ω ⃗ ~ B $Q B A$ T θ = $Q B A$ e ω ⃗ ~ B θ $Q B A$ T ← e P A P − 1 = P e A P − 1 Rot\left( \vec{\omega}^A,\theta \right) =e^{\tilde{\vec{\omega}}^A\theta}=e^{\widetilde{\left $Q_{\\mathrm{B}}\^{A} \\right$ \vec{\omega}^B}\theta}=e^{\left $Q_{\\mathrm{B}}\^{A} \\right$ \tilde{\vec{\omega}}^B\left $Q_{\\mathrm{B}}\^{A} \\right$ ^{\mathrm{T}}\theta}=\left $Q_{\\mathrm{B}}\^{A} \\right$ e^{\tilde{\vec{\omega}}^B\theta}\left $Q_{\\mathrm{B}}\^{A} \\right$ ^{\mathrm{T}}\gets e^{PAP^{-1}}=Pe^AP^{-1} Rot(ω A,θ)=eω ~Aθ=e $QBA$ ω B θ=e $QBA$ ω ~B $QBA$ Tθ= $QBA$ eω ~Bθ $QBA$ T←ePAP−1=PeAP−1

3. Rigid-Body Operation via Diffeential Equation

Recall: Every $Q$ ∈ S O ( 3 ) \left $Q \\right$ \in SO\left( 3 \right) $Q$ ∈SO(3) can be viewed as the state transition matrix associated with the rotation ODE(1). It maps the initial position to the current position(after the rotation motion)

p ⃗ ( θ ) = R o t ( ω ⃗ , θ ) p ⃗ 0 \vec{p}\left( \theta \right) =Rot\left( \vec{\omega},\theta \right) \vec{p}_0 p (θ)=Rot(ω ,θ)p 0 viewed as solution to p ⃗ ˙ ( t ) = ω ⃗ ~ p ⃗ ( t ) , p ⃗ ( 0 ) = p ⃗ 0 \dot{\vec{p}}\left( t \right) =\tilde{\vec{\omega}}\vec{p}\left( t \right) ,\vec{p}\left( 0 \right) =\vec{p}_0 p ˙(t)=ω ~p (t),p (0)=p 0 at t = θ t=\theta t=θ
The above relation requires that the rotation axis passes through the origin.

We can obtain similar ODE characterization for $T$ ∈ S E ( 3 ) \left $T \\right$ \in SE\left( 3 \right) $T$ ∈SE(3) , which will lead to exponential coordinate of SE(3)

Recall : Theorem (Chasles): Every rigid body motion can be realized by a screw motion

Consider a point p p p undergoes a screw motion with screw axis S \mathcal{S} S and unit speed ( θ ˙ = 1 \dot{\theta}=1 θ˙=1) . Let the cooresponding twist be V = θ ˙ S = ( ω ⃗ , v ⃗ ) \mathcal{V} =\dot{\theta}\mathcal{S} =\left( \vec{\omega},\vec{v} \right) V=θ˙S=(ω ,v ) . The motion can be described be the following ODE.
p ⃗ ˙ ( t ) = ω ⃗ × p ⃗ ( t ) + v ⃗ ⇒ $p ⃗ ˙ ( t ) 0$ = $ω ⃗ \~ v ⃗ 0 0$ $p ⃗ ( t ) 1$ \dot{\vec{p}}\left( t \right) =\vec{\omega}\times \vec{p}\left( t \right) +\vec{v}\Rightarrow \left $\\begin{array}{c} \\dot{\\vec{p}}\\left( t \\right)\\\\ 0\\\\ \\end{array} \\right$ =\left $\\begin{matrix} \\tilde{\\vec{\\omega}}\& \\vec{v}\\\\ 0\& 0\\\\ \\end{matrix} \\right$ \left $\\begin{array}{c} \\vec{p}\\left( t \\right)\\\\ 1\\\\ \\end{array} \\right$ p ˙(t)=ω ×p (t)+v ⇒ $p ˙(t)0$ = $ω \~0v 0$ $p (t)1$

Solution in homogeneous coordinate is :

p ⃗ ( t ) 1 \] = exp ⁡ ( \[ ω ⃗ \~ v ⃗ 0 0 \] t ) \[ p ⃗ ( 0 ) 1 \] \\left\[ \\begin{array}{c} \\vec{p}\\left( t \\right)\\\\ 1\\\\ \\end{array} \\right\] =\\exp \\left( \\left\[ \\begin{matrix} \\tilde{\\vec{\\omega}}\& \\vec{v}\\\\ 0\& 0\\\\ \\end{matrix} \\right\] t \\right) \\left\[ \\begin{array}{c} \\vec{p}\\left( 0 \\right)\\\\ 1\\\\ \\end{array} \\right\] \[p (t)1\]=exp(\[ω \~0v 0\]t)\[p (0)1

3.1 SE(3)

For any twist V = ( ω ⃗ , v ⃗ ) \mathcal{V} =\left( \vec{\omega},\vec{v} \right) V=(ω ,v ), let $V$ \left $\\mathcal{V} \\right$ $V$ be its matrix representation of twist V \mathcal{V} V
$V$ = $ω ⃗ \~ v ⃗ 0 0$ ∈ R 4 × 4 \left $\\mathcal{V} \\right$ =\left $\\begin{matrix} \\tilde{\\vec{\\omega}}\& \\vec{v}\\\\ 0\& 0\\\\ \\end{matrix} \\right$ \in \mathbb{R} ^{4\times 4} $V$ = $ω \~0v 0$ ∈R4×4

The abve definition also applies to a screw axis S = ( ω ⃗ , v ⃗ ) , $S$ = $ω ⃗ \~ v ⃗ 0 0$ \mathcal{S} =\left( \vec{\omega},\vec{v} \right) ,\left $\\mathcal{S} \\right$ =\left $\\begin{matrix} \\tilde{\\vec{\\omega}}\& \\vec{v}\\\\ 0\& 0\\\\ \\end{matrix} \\right$ S=(ω ,v ), $S$ = $ω \~0v 0$
With this notation, the solution is $p ⃗ ( t ) 1$ = e $S$ t $p ⃗ ( 0 ) 1$ \left $\\begin{array}{c} \\vec{p}\\left( t \\right)\\\\ 1\\\\ \\end{array} \\right$ =e^{\left $\\mathcal{S} \\right$ t}\left $\\begin{array}{c} \\vec{p}\\left( 0 \\right)\\\\ 1\\\\ \\end{array} \\right$ $p (t)1$ =e $S$ t $p (0)1$
Fact: e $S$ t ∈ S E ( 3 ) e^{\left $\\mathcal{S} \\right$ t}\in SE\left( 3 \right) e $S$ t∈SE(3) is always a valid homogeneous transformation matrix.
$T \] = e \[ S \] t = \[ \[ Q \] R ⃗ 0 1 \] \\left\[ T \\right\] =e\^{\\left\[ \\mathcal{S} \\right\] t}=\\left\[ \\begin{matrix} \\left\[ Q \\right\]\& \\vec{R}\\\\ 0\& 1\\\\ \\end{matrix} \\right\] \[T\]=e\[S\]t=\[\[Q\]0R 1$
Fact: Any T ∈ S E ( 3 ) T\in SE\left( 3 \right) T∈SE(3) can be written as $T$ = e $S$ t \left $T \\right$ =e^{\left $\\mathcal{S} \\right$ t} $T$ =e $S$ t, i.e. , it can be viewed as an operator that moves a point/frame along the screw axis S \mathcal{S} S at unit speed for time t t t

3.2 se(3)

∀ ω ⃗ ∈ R 3 → ω ⃗ ~ ∈ s o ( 3 ) → e ω ⃗ ~ θ ∈ S O ( 3 ) ∀ S ∈ R 6 → $S$ ∣ 4 × 4 = $ω ⃗ \~ v ⃗ 0 0$ ∈ s e ( 3 ) → e $S$ θ ∈ S E ( 3 ) \forall \vec{\omega}\in \mathbb{R} ^3\rightarrow \tilde{\vec{\omega}}\in so\left( 3 \right) \rightarrow e^{\tilde{\vec{\omega}}\theta}\in SO\left( 3 \right) \\ \forall \mathcal{S} \in \mathbb{R} ^6\rightarrow \left. \left $\\mathcal{S} \\right$ \right|_{4\times 4}=\left $\\begin{matrix} \\tilde{\\vec{\\omega}}\& \\vec{v}\\\\ 0\& 0\\\\ \\end{matrix} \\right$ \in se\left( 3 \right) \rightarrow e^{\left $\\mathcal{S} \\right$ \theta}\in SE\left( 3 \right) ∀ω ∈R3→ω ~∈so(3)→eω ~θ∈SO(3)∀S∈R6→ $S$ ∣4×4= $ω \~0v 0$ ∈se(3)→e $S$ θ∈SE(3)

Similar to s o ( 3 ) so\left( 3 \right) so(3) , we can define s e ( 3 ) se\left( 3 \right) se(3) :
s e ( 3 ) = { ( ω ⃗ ~ , v ⃗ ) , ω ⃗ ~ ∈ s o ( 3 ) , v ⃗ ∈ R 3 } se\left( 3 \right) =\left\{ \left( \tilde{\vec{\omega}},\vec{v} \right) ,\tilde{\vec{\omega}}\in so\left( 3 \right) ,\vec{v}\in \mathbb{R} ^3 \right\} se(3)={(ω ~,v ),ω ~∈so(3),v ∈R3}

s e ( 3 ) se\left( 3 \right) se(3) contains all matrix representation of twists or equivalently all twists.
In some references, $V$ \left $\\mathcal{V} \\right$ $V$ is called a twist
Sometimes, we may abuse notation by writing V ∈ s e ( 3 ) \mathcal{V} \in se\left( 3 \right) V∈se(3)

4. Homogeneous Transformation Matrix as Rigid-Body Operator

ODE for rigid motion under V = ( ω ⃗ , v ⃗ ) \mathcal{V} =\left( \vec{\omega},\vec{v} \right) V=(ω ,v )
p ⃗ ˙ = v ⃗ + ω ⃗ × p ⃗ ⇒ $p ⃗ ˙ 0$ = $ω ⃗ \~ v ⃗ 0 0$ $p ⃗ 1$ ⇒ $p ⃗ ˙ 0$ = e $V$ t $p ⃗ 1$ \dot{\vec{p}}=\vec{v}+\vec{\omega}\times \vec{p}\Rightarrow \left $\\begin{array}{c} \\dot{\\vec{p}}\\\\ 0\\\\ \\end{array} \\right$ =\left $\\begin{matrix} \\tilde{\\vec{\\omega}}\& \\vec{v}\\\\ 0\& 0\\\\ \\end{matrix} \\right$ \left $\\begin{array}{c} \\vec{p}\\\\ 1\\\\ \\end{array} \\right$ \Rightarrow \left $\\begin{array}{c} \\dot{\\vec{p}}\\\\ 0\\\\ \\end{array} \\right$ =e^{\left $\\mathcal{V} \\right$ t}\left $\\begin{array}{c} \\vec{p}\\\\ 1\\\\ \\end{array} \\right$ p ˙=v +ω ×p ⇒ $p ˙0$ = $ω \~0v 0$ $p 1$ ⇒ $p ˙0$ =e $V$ t $p 1$
Consider "unit velocity" V = S \mathcal{V} =\mathcal{S} V=S, then time t t t means degree

if not unit speed : V = θ ˙ S \mathcal{V} =\dot{\theta}\mathcal{S} V=θ˙S
$p ⃗ ' 1$ = $T$ $p ⃗ 1$ \left $\\begin{array}{c} \\vec{p}\^{\\prime}\\\\ 1\\\\ \\end{array} \\right$ =\left $T \\right$ \left $\\begin{array}{c} \\vec{p}\\\\ 1\\\\ \\end{array} \\right$ $p '1$ = $T$ $p 1$ : "rotate" p ⃗ \vec{p} p about screw axis S \mathcal{S} S by θ \theta θ degree
$T \] = e \[ S \] θ \\left\[ T \\right\] =e\^{\\left\[ \\mathcal{S} \\right\] \\theta} \[T\]=e\[S\]θ, two points : \[ p ⃗ 1 \] → \[ p ⃗ ′ 1 \] \\left\[ \\begin{array}{c} \\vec{p}\\\\ 1\\\\ \\end{array} \\right\] \\rightarrow \\left\[ \\begin{array}{c} \\vec{p}\^{\\prime}\\\\ 1\\\\ \\end{array} \\right\] \[p 1\]→\[p ′1\] , more precisely : \[ p ⃗ O ′ 1 \] = \[ T O \] \[ p ⃗ O 1 \] \\left\[ \\begin{array}{c} {\\vec{p}\^O}\^{\\prime}\\\\ 1\\\\ \\end{array} \\right\] =\\left\[ T\^O \\right\] \\left\[ \\begin{array}{c} \\vec{p}\^O\\\\ 1\\\\ \\end{array} \\right\] \[p O′1\]=\[TO\]\[p O1$
For $T$ ∈ S E ( 3 ) \left $T \\right$ \in SE\left( 3 \right) $T$ ∈SE(3) config representative $T B A$ \left $T_{\\mathrm{B}}\^{A} \\right$ $TBA$ : config of { B } \left\{ B \right\} {B} relative to { A } \left\{ A \right\} {A} ------ $p ⃗ A 1$ = $T B A$ $p ⃗ B 1$ \left $\\begin{array}{c} \\vec{p}\^A\\\\ 1\\\\ \\end{array} \\right$ =\left $T_{\\mathrm{B}}\^{A} \\right$ \left $\\begin{array}{c} \\vec{p}\^B\\\\ 1\\\\ \\end{array} \\right$ $p A1$ = $TBA$ $p B1$ ------same physical point but two different frames
$T$ $T A$ \left $T \\right$ \left $T_A \\right$ $T$ $TA$ : "rotate" { A } \left\{ A \right\} {A}-frame about S \mathcal{S} S by θ \theta θ degree

Rigid-Body Operator in Different Frames

Expression of $T$ \left $T \\right$ $T$ in another frame (other than { O } \left\{ O \right\} {O}):
$T O \] ↔ \[ T B O \] − 1 \[ T O \] \[ T B O \] \\left\[ T\^O \\right\] \\leftrightarrow \\left\[ T_{\\mathrm{B}}\^{O} \\right\] \^{-1}\\left\[ T\^O \\right\] \\left\[ T_{\\mathrm{B}}\^{O} \\right\] \[TO\]↔\[TBO\]−1\[TO\]\[TBO$

5. Rigid-Body Operation of Screw Axis

Consider an arbitrary screw axis S \mathcal{S} S , suppose the axis has gone through a rigid transformation $T$ = ( $Q$ , R ⃗ ) \left $T \\right$ =\left( \left $Q \\right$ ,\vec{R} \right) $T$ =( $Q$ ,R ) and the resulting new screw axis is S ′ \mathcal{S} ^{\prime} S′ , then
S ′ = $A d T$ S \mathcal{S} ^{\prime}=\left $Ad_T \\right$ \mathcal{S} S′= $AdT$ S

Let's work an arbitrary frame { A } \left\{ A \right\} {A} (rigidly attached to the screw axis)

Let { B } \left\{ B \right\} {B} be the frame obtained be apply $T$ \left $T \\right$ $T$ operation

the coordinate of S \mathcal{S} S in { A } \left\{ A \right\} {A} is the same as the coordinate of S ′ \mathcal{S} ^{\prime} S′ in { B } \left\{ B \right\} {B} : S A = S ′ B \mathcal{S} ^A={\mathcal{S} ^{\prime}}^B SA=S′B

We also know $T B$ = $T$ $T A$ , $T$ = $T B A$ \left $T_B \\right$ =\left $T \\right$ \left $T_{\\mathrm{A}} \\right$ ,\left $T \\right$ =\left $T_{\\mathrm{B}}\^{A} \\right$ $TB$ = $T$ $TA$ , $T$ = $TBA$

Multiply $X B A$ \left $X_{\\mathrm{B}}\^{A} \\right$ $XBA$ : ⇒ $X B A$ S A = $X B A$ S ′ B = S ′ A ⇒ S ′ A = $X B A$ S A \Rightarrow \left $X_{\\mathrm{B}}\^{A} \\right$ \mathcal{S} ^A=\left $X_{\\mathrm{B}}\^{A} \\right$ {\mathcal{S} ^{\prime}}^B={\mathcal{S} ^{\prime}}^A\Rightarrow {\mathcal{S} ^{\prime}}^A=\left $X_{\\mathrm{B}}\^{A} \\right$ \mathcal{S} ^A ⇒ $XBA$ SA= $XBA$ S′B=S′A⇒S′A= $XBA$ SA
$X B A \] = \[ \[ Q B A \] 0 R ⃗ \~ B A \[ Q B A \] \[ Q B A \] \] = \[ A d T \] \\left\[ X_{\\mathrm{B}}\^{A} \\right\] =\\left\[ \\begin{matrix} \\left\[ Q_{\\mathrm{B}}\^{A} \\right\]\& 0\\\\ \\tilde{\\vec{R}}_{\\mathrm{B}}\^{A}\\left\[ Q_{\\mathrm{B}}\^{A} \\right\]\& \\left\[ Q_{\\mathrm{B}}\^{A} \\right\]\\\\ \\end{matrix} \\right\] =\\left\[ Ad_T \\right\] \[XBA\]=\[\[QBA\]R \~BA\[QBA\]0\[QBA\]\]=\[AdT$